Question

# Suppose you ask a friend to randomly choose an integer between 1 and 10, inclusive. What is the probability that the number will be more than 4 or odd

Equation, expression, and inequalitie
Suppose you ask a friend to randomly choose an integer between 1 and 10, inclusive. What is the probability that the number will be more than 4 or odd? (Enter your probability as a fraction.)

2021-06-25

6 of the 10 integers between 1 and 10 inclusive are more than 4 (that is, 5,6,7,8,9,10).
The probability is the number of favorable outcomes divided by the number of possible outcomes: $$P(>4)=\#$$of favorable outcomes $$\#$$ of possible outcomes=$$\frac{6}{10}$$
5 of the 10 integers between 1 and 10 inclusive are odd (that is, 1,3,5,7,9). P(odd)=# of favorable outcomes/# of possible outcomes=$$\frac{5}{10}$$
3 of the 10 integers between 1 and 10 inclusive are move than 4 and odd (that is, 5,7,9). $$P(>4$$ and odd$$)=\#$$ of favorable outcomes/# of possible outcomes=$$\frac{3}{10}$$
Use the General addition rule for any two events: $$P(AUB)=P(A)+P(B)-P(A⋂B)$$
P(>4 or odd) $$=P(>4)+P(odd)-P(>4 \text{and odd}) =\frac{6}{10}+\frac{5}{10}-\frac{3}{10} =\frac{6+5-3}{10} =\frac{8}{10} =\frac{4}{5}$$